Since \(A/(x^{2c+1})\cong A^{\prime }/(x^{2c+1})\) , \(y^{\prime }\in A^n\)  can be choose such that \(v(Y)=y^{\prime } + (x^{2c+1})\) .
Set \(P=LM\)  and \(d=P(y^{\prime })\) . We have \(dA=x^cA\) .
Let \(H\)  be the \(n\times n\) -matrix obtained adding down to \((\partial f/\partial Y)\)  as a border the block \((0|\mbox{Id}_{n-r})\) . Let \(G^{\prime }\)  be the adjoint matrix of \(H\)  and \(G=LG^{\prime }\) . We have
\(GH=HG=LM \mbox{Id}_n=P\mbox{Id}_n\) 
and so
\(d\mbox{Id}_n=P(y^{\prime })\mbox{Id}_n=G(y^{\prime })H(y^{\prime }).\)
Let
\(h=Y-y^{\prime }-dG(y^{\prime })T,\) 
where \(T=(T_1,\ldots ,T_n)\)  are new variables. Since
\(Y-y^{\prime }\equiv dG(y^{\prime })T\ \mbox{modulo}\ h\) 
and
\(f(Y)-f(y^{\prime })\equiv \sum _j\partial f/\partial Y_j(y^{\prime }) (Y_j-y^{\prime }_j)\) 
modulo higher order terms in \(Y_j-y^{\prime }_j\) , by Taylor's formula we see that we have
\(f(Y)-f(y^{\prime })\equiv \sum _jd\partial f/\partial Y_j(y^{\prime }) G_j(y^{\prime })T_j+d^2Q=dP(y^{\prime })T+d^2Q=d^2(T+Q)\) 
\(\mbox{modulo}\ h,\) 
where \(Q\in T^2 A[T]^r\) . This is because \((\partial f/\partial Y)G=(P\mbox{Id}_r|0)\) . We have \(f(y^{\prime })=d^2a\)  for some \(a\in xA^r\) .
Set \(g_i=a_i+T_i+Q_i\) , \(i\in [r]\)  and \(E=A[Y,T]/(I,g,h)\) .
Then there exists \(s,s^{\prime }\)  in \(1+(T)\)  as it is proved in [1]} such that \(C:=E_{ss^{\prime }}\cong (A[T]/(g))_{ss^{\prime }}\)  is smooth . Moreover, \(v^{\prime }\)  factors through \(C \) .

